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Algebra - Factorization of polynomials (Remainder and Factor Theorems)

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

๐Ÿ”‘Concepts

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Definition of a Polynomial: A polynomial f(x)f(x) in one variable xx is an algebraic expression of the form anxn+anโˆ’1xnโˆ’1+...+a1x+a0a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0. Visually, the graph of a polynomial is a smooth, continuous curve without any breaks or sharp corners, where the degree of the polynomial determines the maximum number of times the curve can turn.

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The Division Algorithm: For any polynomial f(x)f(x) and a divisor g(x)g(x), there exist unique polynomials q(x)q(x) (quotient) and r(x)r(x) (remainder) such that f(x)=g(x)โ‹…q(x)+r(x)f(x) = g(x) \cdot q(x) + r(x), where the degree of r(x)r(x) is always less than the degree of g(x)g(x).

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The Remainder Theorem: If a polynomial f(x)f(x) is divided by (xโˆ’a)(x - a), the remainder is equal to f(a)f(a). Visually, this remainder represents the yy-coordinate of the point on the graph of y=f(x)y = f(x) where the xx-coordinate is aa. If the remainder is positive, the point is above the xx-axis; if negative, it is below.

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The Factor Theorem: A polynomial (xโˆ’a)(x - a) is a factor of f(x)f(x) if and only if f(a)=0f(a) = 0. In graphical terms, if (xโˆ’a)(x - a) is a factor, the curve y=f(x)y = f(x) will cross or touch the xx-axis at the point (a,0)(a, 0), making x=ax = a an xx-intercept of the graph.

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General Remainder Rule: When a polynomial f(x)f(x) is divided by (axโˆ’b)(ax - b), the remainder is f(ba)f(\frac{b}{a}). If divided by (ax+b)(ax + b), the remainder is f(โˆ’ba)f(-\frac{b}{a}). This is found by setting the linear divisor to zero (axโˆ’b=0ax - b = 0) and solving for xx.

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Complete Factorization: To factorize a cubic polynomial ax3+bx2+cx+dax^3 + bx^2 + cx + d, we first find one factor (xโˆ’k)(x - k) using the Factor Theorem (by testing factors of the constant term dd). We then use synthetic division or long division to divide the polynomial by (xโˆ’k)(x - k) to obtain a quadratic quotient, which is then factorized further into two linear factors if possible.

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Multiple Factors and Simultaneous Equations: If a polynomial has two unknown constants (e.g., aa and bb) and we are given two factors or two remainders, we can substitute the values into f(x)f(x) to form a system of two linear equations. Solving these equations simultaneously reveals the values of aa and bb.

๐Ÿ“Formulae

f(x)=(xโˆ’a)Q(x)+Rf(x) = (x - a)Q(x) + R

Remainder R=f(a)R = f(a) when f(x)f(x) is divided by (xโˆ’a)(x - a)

Remainder R=f(โˆ’ba)R = f(-\frac{b}{a}) when f(x)f(x) is divided by (ax+b)(ax + b)

If f(a)=0โ€…โ€ŠโŸนโ€…โ€Š(xโˆ’a)f(a) = 0 \implies (x - a) is a factor of f(x)f(x)

If (xโˆ’a)(x - a) and (xโˆ’b)(x - b) are factors, then (xโˆ’a)(xโˆ’b)(x - a)(x - b) is also a factor of f(x)f(x)

๐Ÿ’กExamples

Problem 1:

Find the remainder when the polynomial f(x)=2x3โˆ’5x2+xโˆ’7f(x) = 2x^3 - 5x^2 + x - 7 is divided by (xโˆ’3)(x - 3).

Solution:

  1. Let f(x)=2x3โˆ’5x2+xโˆ’7f(x) = 2x^3 - 5x^2 + x - 7.
  2. The divisor is (xโˆ’3)(x - 3). According to the Remainder Theorem, the remainder R=f(3)R = f(3).
  3. Substitute x=3x = 3 into the polynomial: f(3)=2(3)3โˆ’5(3)2+(3)โˆ’7f(3) = 2(3)^3 - 5(3)^2 + (3) - 7
  4. Calculate the powers: f(3)=2(27)โˆ’5(9)+3โˆ’7f(3) = 2(27) - 5(9) + 3 - 7
  5. Multiply and simplify: f(3)=54โˆ’45+3โˆ’7f(3) = 54 - 45 + 3 - 7
  6. f(3)=9+3โˆ’7=5f(3) = 9 + 3 - 7 = 5. Therefore, the remainder is 55.

Explanation:

The Remainder Theorem allows us to find the remainder without performing long division. We simply evaluate the polynomial at the value that makes the divisor zero.

Problem 2:

Show that (xโˆ’2)(x - 2) is a factor of x3โˆ’3x2+4xโˆ’4x^3 - 3x^2 + 4x - 4. Hence, factorize the polynomial completely.

Solution:

  1. Let f(x)=x3โˆ’3x2+4xโˆ’4f(x) = x^3 - 3x^2 + 4x - 4. To show (xโˆ’2)(x - 2) is a factor, we must prove f(2)=0f(2) = 0.
  2. f(2)=(2)3โˆ’3(2)2+4(2)โˆ’4=8โˆ’12+8โˆ’4=16โˆ’16=0f(2) = (2)^3 - 3(2)^2 + 4(2) - 4 = 8 - 12 + 8 - 4 = 16 - 16 = 0. Since f(2)=0f(2) = 0, (xโˆ’2)(x - 2) is a factor.
  3. Divide f(x)f(x) by (xโˆ’2)(x - 2) using long division: (x3โˆ’3x2+4xโˆ’4)รท(xโˆ’2)=x2โˆ’x+2(x^3 - 3x^2 + 4x - 4) \div (x - 2) = x^2 - x + 2.
  4. The polynomial can be written as f(x)=(xโˆ’2)(x2โˆ’x+2)f(x) = (x - 2)(x^2 - x + 2).
  5. Check if the quadratic x2โˆ’x+2x^2 - x + 2 can be factorized further. The discriminant D=b2โˆ’4ac=(โˆ’1)2โˆ’4(1)(2)=1โˆ’8=โˆ’7D = b^2 - 4ac = (-1)^2 - 4(1)(2) = 1 - 8 = -7. Since D<0D < 0, the quadratic has no real factors.
  6. Final factorization: (xโˆ’2)(x2โˆ’x+2)(x - 2)(x^2 - x + 2).

Explanation:

First, the Factor Theorem is used to verify the given factor. Then, division is performed to reduce the cubic polynomial to a quadratic, which is then checked for further factorization.